Answer
$i=\frac{1}{5}[-4e^{-4t}-12\sin (3t)+9 \cos (3t)]$
Work Step by Step
Using RC Circuit we get:
$$\frac{dq}{dt}+\frac{q}{RC}=\frac{E}{R}$$
We are given:
$E=10\cos 3t$
$C=\frac{1}{8}R$
$R=2$
Substituting:
$$\frac{dq}{dt}+4q=5\cos 3t$$
The solution can be found by:
$$Q=e^{-\int4 dt}(c+\int 5\cos (3t)e^{4}dt)$$
$$Q=ce^{-4t}+e^{-4t}[\frac{1}{5})3\sin (3t)e^{4t}+4\cos(3t))e^{4t}]$$
$$Q=ce^{-4t}+\frac{1}{5}[3\sin (3t)+4\cos(3t)]$$
Since $q(0)=1$, we get:
$$1=c+\frac{4}{5} \rightarrow c=\frac{1}{5}$$
The charge is:
$$q=\frac{1}{5}e^{-4t}+\frac{1}{5}[3\sin (3t)+4\cos (3t)]$$
Thus the current in the circuit for $t \geq 0$ is:
$$i=\frac{1}{5}[-4e^{-4t}-12\sin (3t)+9 \cos (3t)]$$
